Đáp án:
Theo Pytago ta có:
$\begin{array}{l}
a)B{C^2} = A{B^2} + A{C^2}\\
\Rightarrow BC = \sqrt {A{B^2} + A{C^2}} = \sqrt {{3^2} + {3^2}} = 3\sqrt 2 \left( {cm} \right)\\
b)BC = \sqrt {{4^2} + {6^2}} = \sqrt {52} = 2\sqrt {13} \left( {cm} \right)\\
c)BC = \sqrt {2,{3^2} + 3,{9^2}} = \frac{{\sqrt {82} }}{2}\left( {cm} \right)\\
d)BC = \sqrt {{{\left( {\sqrt 5 } \right)}^2} + {3^2}} = \sqrt {14} \left( {cm} \right)\\
e)BC = \sqrt {8 + 17} = \sqrt {25} = 5\left( {cm} \right)
\end{array}$
